A125162 a(n) is the number of primes of the form k! + n, 1 <= k <= n.

1, 1, 1, 1, 3, 1, 4, 0, 1, 1, 5, 1, 3, 0, 1, 1, 6, 1, 7, 0, 1, 1, 6, 0, 1, 0, 1, 1, 6, 1, 9, 0, 0, 0, 3, 1, 11, 0, 1, 1, 9, 1, 5, 0, 1, 1, 10, 0, 2, 0, 1, 1, 9, 0, 2, 0, 1, 1, 10, 1, 9, 0, 0, 0, 3, 1, 9, 0, 1, 1, 8, 1, 9, 0, 0, 0, 5, 1, 9, 0, 1, 1, 11, 0, 1, 0, 1, 1, 8, 0, 3, 0, 0, 0, 2, 1, 10, 0, 1, 1, 10, 1

Offset: 1

Author

Alexander Adamchuk, Nov 21 2006

Keywords

• nonn

Comments

Note the triples of consecutive zeros in a(n) for n = {{32,33,34}, {62,63,64}, {74,75,76}, {92,93,94}, {116,117,118}, {122,123,124}, {140,141,142}, {152,153,154}, {158,159,160}, {182,183,184}, {200,201,202}, {206,207,208}, {212,213,214}, {218,219,220}, {242,243,244}, {272,273,274}, {284,285,286}, ...}. The middle index of most zero triples is a multiple of 3. See A125164.

The first consecutive quintuple of zeros has indices n = {294,295,296,297,298}, where the odd zero index n = 295 is not a multiple of 3.

Also for n >= 2, a(n) is the number of primes of the form k! + n for all k, since n divides k! + n for k >= n. Note that it is not known whether there are infinitely many primes of the form k! + 1; see A088332 for such primes and A002981 for the indices k. - Jianing Song, Jul 28 2018

Examples

			a(n) is the length of n-th row in the table of numbers k such that k! + n is a prime, 1 <= k <= n.
   n:  numbers k
   -------------
   1:  {1},
   2:  {1},
   3:  {2},
   4:  {1},
   5:  {2, 3, 4},
Thus a(1)-a(4) = 1, a(5) = 3.
See Example table link for more rows.
		

Links

• Michel Marcus, Example table

Crossrefs

Cf. A125163 (indices of 0), A125164 (triples).

Programs

Mathematica

Table[Length[Select[Range[n],PrimeQ[ #!+n]&]],{n,1,300}]

PARI

a(n)=c=0;for(k=1,n,if(ispseudoprime(k!+n),c++));c
vector(100,n,a(n)) \\ Derek Orr, Oct 15 2014

Extensions

Name clarified by Jianing Song, Jul 28 2018

Edited by Michel Marcus, Jul 29 2018